$\bf \sqrt{48n^9}\qquad\begin{cases}
48=2\cdot 2\cdot 2\cdot 2\cdot 3\\
\qquad 2^2\cdot 2^2\cdot 3\\
\qquad (2^2)^2\cdot 3\\
n^9=n^{8+1}\\
\qquad n^8\cdot n^1\\
\qquad n^{4\cdot 2}\cdot n\\
\qquad (n^4)^2\cdot n
\end{cases}\implies \sqrt{(2^2)^2\cdot 3\cdot (n^4)^2\cdot n}
\\\\\\
2^2n^4\sqrt{3n}\implies 4n^4\sqrt{3n}$

3 0

4 years ago

HELP PLEASE!!!!!!!!!!<br><br> (btw the other screen shot is what i put for b)

White raven [17]

Answer:

Step-by-step explanation:

took the test

5 0

3 years ago

Read 2 more answers

How do i express x^2-3x into the form of (x-m)^2+n.

Kipish [7]

Well this could be incredibly easy, by just saying $m=0$ and $n=-3x$ .

This is a completely valid case but nevertheless seems too easy. The requirements are probably that n and m are integers or natural numbers but since that was not specified such probable requirement ought not to be followed.

Hope this helps :)

6 0

3 years ago

Which reflection of the point will produce an image at the same time coordinates (0,k)

Anettt [7]

I need more details.

4 0

3 years ago